Question

The expression $\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}}$ can be written as:
A. ${\text{secAcosecA + 1}}$
B. tan A + cot A
C. sec A + cosec A.
D. sin A cos A + 1

Answer 1

Hint: To solve this question, we will use some basic trigonometric identities to simplify the given expression. Some useful identities are: $\cot A = \dfrac{1}{{\tan A}}$, $\cot A = \dfrac{{\cos A}}{{\sin A}}$, $\tan A = \dfrac{{\sin A}}{{\cos A}}$.

Complete step-by-step answer:
We have,
$ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}}$ ……… (i)
Replacing $\cot A = \dfrac{1}{{\tan A}}$ in equation (i), we will get
$
   \Rightarrow \dfrac{{\tan A}}{{1 - \dfrac{1}{{\tan A}}}} + \dfrac{{\dfrac{1}{{\tan A}}}}{{1 - \tan A}} \\
   \Rightarrow \dfrac{{{{\tan }^2}A}}{{\tan A - 1}} + \dfrac{1}{{\tan A\left( {1 - \tan A} \right)}} \\
   \Rightarrow \dfrac{{{{\tan }^2}A}}{{\tan A - 1}} - \dfrac{1}{{\tan A\left( {\tan A - 1} \right)}} \\
$
Adding by taking L.C.M $\tan A\left( {\tan A - 1} \right)$
$ \Rightarrow \dfrac{{{{\tan }^3}A - 1}}{{\tan A\left( {\tan A - 1} \right)}}$
Now, expanding ${\tan ^3}A - 1$ by using the identity ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$
We will get,
 \[ \Rightarrow \dfrac{{\left( {\tan A - 1} \right)\left( {{{\tan }^2}A + 1 + \tan A} \right)}}{{\tan A\left( {\tan A - 1} \right)}}\]
Dividing numerator and denominator by $\left( {\tan A - 1} \right)$,
\[ \Rightarrow \dfrac{{\left( {{{\tan }^2}A + 1 + \tan A} \right)}}{{\tan A}}\]
Solving this,
\[ \Rightarrow \tan A + \cot A + 1\]
Now, replacing $\cot A = \dfrac{{\cos A}}{{\sin A}}$ and $\tan A = \dfrac{{\sin A}}{{\cos A}}$
\[ \Rightarrow \dfrac{{\sin A}}{{\cos A}} + \dfrac{{\cos A}}{{\sin A}} + 1\]
Adding by taking L.C.M,
\[ \Rightarrow \dfrac{{{{\sin }^2}A + {{\cos }^2}A}}{{\cos A\sin A}} + 1\]
As we know that,
\[{\sin ^2}A + {\cos ^2}A = 1\]
Putting this, we will get
\[
   \Rightarrow \dfrac{1}{{\cos A\sin A}} + 1 \\
   \Rightarrow \dfrac{1}{{\cos A}} \times \dfrac{1}{{\sin A}} + 1 \\
   \Rightarrow {\text{secAcosecA}} + 1 \\
\]
Hence, the expression $\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}}$ can be written as secAcosecA + 1.
Therefore, the correct answer is option(A).

Note: This question can also be solved by another method using the identities of sin A and cos A. The solution is as follows:
$
   \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} \\
   \Rightarrow \dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{1 - \dfrac{{\cos A}}{{\sin A}}}} + \dfrac{{\dfrac{{\cos A}}{{\sin A}}}}{{1 - \dfrac{{\sin A}}{{\cos A}}}} \\
    \\
$
$
   \Rightarrow \dfrac{{{{\sin }^2}A}}{{\cos A\left( {\sin A - \cos A} \right)}} - \dfrac{{{{\cos }^2}A}}{{\sin A\left( {\sin A - \cos A} \right)}} \\
   \Rightarrow \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{\sin A\cos A\left( {\sin A - \cos A} \right)}} \\
   \Rightarrow \dfrac{{\left( {\sin A - \cos A} \right)\left( {{{\sin }^2}A + {{\cos }^2}A + \sin A\cos A} \right)}}{{\sin A\cos A\left( {\sin A - \cos A} \right)}} \\
   \Rightarrow \dfrac{{\left( {1 + \sin A\cos A} \right)}}{{\sin A\cos A}} \\
   \Rightarrow \dfrac{1}{{\sin A\cos A}} + 1 \\
   \Rightarrow {\text{secAcosecA + 1}} \\
$